📚 IB & OCR Biology: Worked Examples Explained | IB & OCR 生物：典型例题详解
Mastering IB and OCR Biology requires more than memorising facts — it demands the ability to apply concepts to unfamiliar scenarios and structure answers precisely according to mark schemes. This article walks you through ten typical exam-style questions that span cell biology, biochemistry, genetics, evolution and physiology. Each worked example is broken down step by step in both English and Chinese, highlighting key terms, common pitfalls and effective answering strategies.
学好 IB 和 OCR 生物，仅靠记忆知识点远远不够——你需要将概念灵活运用到陌生情境中，并严格按照评分标准组织答案。本文精选十个典型考题，覆盖细胞生物学、生物化学、遗传学、进化和生理学，逐一深度解析。每个例题均采用中英双语逐步拆解，突出关键词、常见错误和高效答题策略。
1. Fluid Mosaic Model & Membrane Transport | 流动镶嵌模型与膜运输
Question: Explain how the structure of the cell membrane is related to its function in controlling the movement of substances into and out of the cell. (6 marks)
题目：解释细胞膜的结构如何与其控制物质进出细胞的功能相关。(6分)
Answer & Commentary:
The phospholipid bilayer forms a hydrophobic core that prevents the free passage of ions and large polar molecules, thereby acting as a selective barrier.
磷脂双层形成疏水核心，阻止离子和大极性分子自由通过，起到选择屏障的作用。
Channel proteins and carrier proteins are embedded in the bilayer. Channel proteins provide hydrophilic pores for facilitated diffusion of specific ions (e.g. Na⁺, K⁺), while carrier proteins change shape to transport molecules such as glucose.
通道蛋白和载体蛋白嵌入双层中。通道蛋白提供亲水孔道，帮助特定离子（如 Na⁺、K⁺）进行协助扩散；载体蛋白则通过构象变化运输葡萄糖等分子。
Cholesterol molecules fit between phospholipids, modulating membrane fluidity and reducing permeability to very small water‑soluble molecules.
胆固醇分子分布在磷脂之间，调节膜的流动性并降低对极小的水溶性分子的通透性。
Glycoproteins and glycolipids on the outer surface form the glycocalyx, which is involved in cell recognition and adhesion, indirectly influencing what enters the cell.
外表面的糖蛋白和糖脂构成糖被，参与细胞识别和黏附，间接影响物质进入。
Active transport is carried out by specific pump proteins (e.g. Na⁺/K⁺‑ATPase) that use energy from ATP to move substances against their concentration gradient, maintaining internal conditions.
主动运输由特定的泵蛋白（如 Na⁺/K⁺‑ATP 酶）完成，利用 ATP 的能量逆浓度梯度转运物质，维持细胞内部环境。
Common mistake: students often forget to link structure explicitly to function, merely listing components. Always state how each component regulates movement.
常见错误：学生常忽略将结构明确与功能联系起来，只罗列成分。务必说明每种成分如何调控物质运动。
2. Enzyme Kinetics & Inhibition | 酶动力学与抑制作用
Question: Describe how competitive and non‑competitive inhibitors affect the rate of an enzyme‑catalysed reaction, using the Michaelis‑Menten model to support your explanation. (5 marks)
题目：运用米氏模型，描述竞争性抑制剂和非竞争性抑制剂如何影响酶促反应速率。(5分)
Answer & Commentary:
A competitive inhibitor has a shape similar to the substrate and binds to the active site, preventing substrate binding. This effect can be overcome by increasing substrate concentration, so Vₘₐₓ remains unchanged but Kₘ increases.
竞争性抑制剂与底物形状相似，能结合活性位点，阻断底物结合。增加底物浓度可克服此抑制，因此 Vₘₐₓ 不变，但 Kₘ 增大。
A non‑competitive inhibitor binds to an allosteric site, altering the enzyme’s conformation so the active site no longer catalyses the reaction effectively. This lowers the number of functional enzyme molecules, reducing Vₘₐₓ, while Kₘ stays the same because the remaining active sites still have normal affinity for the substrate.
非竞争性抑制剂结合别构位点，改变酶构象，使活性位点催化效率下降。功能性酶分子数减少，导致 Vₘₐₓ 降低，而 Kₘ 不变，因为剩余活性位点对底物的亲和力正常。
On a Lineweaver‑Burk plot, competitive inhibition shows lines intersecting on the 1/Vₘₐₓ axis (same y‑intercept), whereas non‑competitive inhibition shows lines intersecting on the −1/Kₘ axis (same x‑intercept).
在 Lineweaver‑Burk 双倒数图中，竞争性抑制的各直线在纵轴截距相同（1/Vₘₐₓ 不变）；非竞争性抑制的各直线在横轴截距相同（−1/Kₘ 不变）。
Key tip: always distinguish between Vₘₐₓ and Kₘ effects and, if required, sketch a graph. Remember that Vₘₐₓ represents the maximum rate when enzyme is saturated.
关键提示：务必区分对 Vₘₐₓ 和 Kₘ 的不同影响，若题目要求，可绘制示意图。牢记 Vₘₐₓ 是酶被底物饱和时的最大速率。
3. Mitosis & the Cell Cycle | 有丝分裂与细胞周期
Question: Outline the stages of mitosis and explain how mitosis ensures genetic consistency between daughter cells. (5 marks)
题目：概述有丝分裂各阶段，并解释有丝分裂如何确保子细胞间的遗传一致性。(5分)
Answer & Commentary:
Prophase: chromosomes condense and become visible as two sister chromatids held by a centromere; the nuclear envelope begins to break down and spindle fibres form.
前期：染色体凝缩，可见由着丝粒连接的两条姐妹染色单体；核膜开始解体，纺锤体形成。
Metaphase: chromosomes line up along the metaphase plate, attached to spindle fibres via kinetochores at the centromeres. This alignment ensures each daughter will receive one copy of each chromatid.
中期：染色体排列在赤道板上，通过着丝粒处的动粒与纺锤丝相连。这一排列确保每个子细胞获得每条染色单体的一个拷贝。
Anaphase: sister chromatids separate at the centromere and are pulled to opposite poles by shortening spindle fibres. The separated chromatids are now individual chromosomes.
后期：姐妹染色单体在着丝粒处分离，由缩短的纺锤丝拉向两极。分离后的染色单体成为独立染色体。
Telophase: chromosomes decondense, nuclear envelopes reform around each set, and cytokinesis divides the cytoplasm, producing two genetically identical daughter cells.
末期：染色体解旋，核膜围绕每组染色体重新形成，胞质分裂完成，产生两个遗传上完全相同的子细胞。
Genetic consistency is maintained because DNA replicates during S phase to produce identical sister chromatids, and anaphase separation ensures each daughter receives an exact copy of the genome.
遗传一致性得以维持是因为 S 期 DNA 复制产生了相同的姐妹染色单体，后期分离确保每个子细胞得到基因组的精确拷贝。
4. DNA Replication | DNA 复制
Question: Explain the role of enzymes in the semi‑conservative replication of DNA. (6 marks)
题目：解释酶在半保留 DNA 复制中的作用。(6分)
Answer & Commentary:
DNA helicase unwinds the double helix by breaking hydrogen bonds between base pairs, forming a replication fork with two template strands.
DNA 解旋酶通过断开碱基对间氢键，解开双螺旋，形成带有两条模板链的复制叉。
Single‑strand binding proteins stabilise the separated strands, preventing them from re‑annealing.
单链结合蛋白稳定已解开的单链，防止它们重新配对。
DNA primase synthesises short RNA primers on each template strand, providing a free 3’‑OH group for DNA polymerase to begin nucleotide addition.
DNA 引物酶在每条模板链上合成短 RNA 引物，提供游离的 3’‑OH 末端，供 DNA 聚合酶开始添加核苷酸。
DNA polymerase III adds free DNA nucleotides to the 3′ end of the primer, synthesising the new strand in the 5’→3′ direction. The leading strand is synthesised continuously; the lagging strand is synthesised discontinuously as Okazaki fragments.
DNA 聚合酶 III 将游离脱氧核苷酸添加到引物的 3′ 端，沿 5’→3′ 方向合成新链。前导链连续合成；后随链不连续合成，形成冈崎片段。
DNA polymerase I removes RNA primers and replaces them with DNA nucleotides. DNA ligase seals the nicks between Okazaki fragments by forming phosphodiester bonds, producing two complete double‑stranded DNA molecules.
DNA 聚合酶 I 切除 RNA 引物并替换为脱氧核苷酸。DNA 连接酶通过形成磷酸二酯键连接冈崎片段间的缺口，最终产生两个完整的双链 DNA 分子。
The process is semi‑conservative because each new DNA molecule consists of one original (parental) strand and one newly synthesised strand, proven by Meselson and Stahl’s experiment using ¹⁵N labelling.
该过程为半保留复制，因为每个新 DNA 分子含有一条旧链（亲本）和一条新合成链，这由 Meselson 和 Stahl 的 ¹⁵N 标记实验证实。
5. Transcription & Translation | 转录与翻译
Question: Describe the process of protein synthesis from DNA to a polypeptide chain. (8 marks)
题目：描述从 DNA 到多肽链的蛋白质合成过程。(8分)
Answer & Commentary:
Transcription: RNA polymerase binds to the promoter region of a gene and unwinds the DNA. It uses one strand as a template to synthesise a complementary pre‑mRNA molecule in the 5’→3′ direction, replacing thymine with uracil.
转录：RNA 聚合酶与基因的启动子区域结合并解开 DNA。它以一条链为模板，沿 5’→3′ 方向合成互补的前体 mRNA，胸腺嘧啶被尿嘧啶代替。
In eukaryotes, pre‑mRNA undergoes splicing: introns are removed and exons are joined to form mature mRNA. A 5′ cap and poly‑A tail are added for stability and nuclear export.
在真核生物中，前体 mRNA 经过剪接：内含子被切除，外显子连接形成成熟 mRNA。同时添加 5′ 帽和 poly‑A 尾以增强稳定性并协助出核。
Translation: the mature mRNA binds to a ribosome. Each codon (triplet of bases) is recognised by a specific tRNA molecule carrying a complementary anticodon and the corresponding amino acid.
翻译：成熟 mRNA 与核糖体结合。每个密码子（三个碱基）被特定的 tRNA 识别，tRNA 带有互补的反密码子和相应的氨基酸。
The ribosome moves along the mRNA; tRNA molecules bring amino acids that are joined by peptide bonds, forming a growing polypeptide chain. This continues until a stop codon is reached, triggering release of the completed polypeptide.
核糖体沿 mRNA 移动；tRNA 逐一搬运氨基酸，氨基酸之间形成肽键，延伸多肽链。直至遇到终止密码子，完整多肽链随即释放。
Post‑translational modifications, such as folding assisted by chaperones, cleavage, or addition of carbohydrate groups, may be required for the protein to become functional.
翻译后修饰，如分子伴侣辅助折叠、剪切或添加糖基，对于蛋白质获得功能往往是必需的。
6. Monohybrid Cross & Pedigree Analysis | 单基因杂交与系谱分析
Question: In a species of plant, red flower colour (R) is dominant to white (r). A heterozygous red‑flowered plant is crossed with a white‑flowered plant. Determine the genotypic and phenotypic ratios of the offspring. Explain what a test cross reveals. (5 marks)
题目：某种植物中，红花 (R) 对白花 (r) 为显性。一株杂合红花植株与白花植株杂交。确定后代的基因型比和表型比，并解释测交能揭示什么信息。(5分)
Answer & Commentary:
Parental genotypes: Rr (red) × rr (white). The gametes produced are R and r from the red parent, and all r from the white parent.
亲本基因型：Rr（红） × rr（白）。红花亲本产生的配子为 R 和 r，白花亲本全部产生 r 配子。
Offspring genotypes: 50% Rr and 50% rr. Thus, the genotypic ratio is 1 Rr : 1 rr. Phenotypically, Rr plants are red, rr plants are white, giving a 1 red : 1 white ratio.
后代基因型：50% Rr、50% rr，基因型比为 1 Rr : 1 rr。表型上，Rr 开红花，rr 开白花，所以表型比为 1 红花 : 1 白花。
A test cross involves crossing an individual showing the dominant phenotype with a homozygous recessive individual. If any offspring show the recessive trait, the unknown parent must be heterozygous; if all offspring show the dominant trait, the parent is likely homozygous dominant. It thus determines the unknown genotype.
测交是将表现显性性状的个体与隐性纯合个体杂交。若后代出现隐性性状，则未知亲本必为杂合；若后代全为显性，则未知亲本很可能为显性纯合。因此测交用于判断个体的未知基因型。
7. Natural Selection & Antibiotic Resistance | 自然选择与抗生素抗性
Question: Explain how the widespread use of antibiotics has led to the evolution of antibiotic‑resistant bacteria, using Darwin’s theory of natural selection. (5 marks)
题目：运用达尔文自然选择学说，解释抗生素的广泛使用如何导致耐药细菌的进化。(5分)
Answer & Commentary:
Within a bacterial population, there is genetic variation due to random mutations. Some mutations confer resistance to a particular antibiotic.
细菌种群中存在遗传变异，源于随机突变。某些突变使细菌获得对特定抗生素的抗性。
When the antibiotic is applied, it acts as a strong selection pressure. Sensitive bacteria are killed or inhibited, while resistant bacteria survive and reproduce.
使用抗生素时，抗生素成为强大的选择压力。敏感细菌被杀死或抑制，而耐药细菌存活并繁殖。
The resistant bacteria pass their resistance alleles to their offspring through vertical gene transfer. Horizontal gene transfer via conjugation, transformation or transduction can also spread resistance genes among different bacteria.
耐药细菌通过垂直基因传递将抗性等位基因传给后代。通过接合、转化或转导进行的水平基因转移也可在不同细菌间传播抗性基因。
Over many generations, the frequency of resistance alleles in the population increases. This is evolution by natural selection — the environment ‘selects’ for the resistant variants.
经过多代，种群中抗性等位基因的频率上升。这就是自然选择驱动的进化——环境“选择”了耐药变种。
To slow resistance, it is essential to complete prescribed antibiotic courses and avoid overuse, reducing the selection pressure that favours resistant strains.
为减缓抗性扩散，须完成全程抗生素治疗并避免滥用，降低对耐药菌株的选择压力。
8. Energy Flow & Trophic Levels | 能量流动与营养级
Question: Explain why the amount of energy available decreases at each successive trophic level in a food chain. (4 marks)
题目：解释为什么食物链中每个连续营养级的可利用能量逐渐减少。(4分)
Answer & Commentary:
Not all of the biomass at one trophic level is consumed by the next level; some material is indigestible and is lost as faeces.
前一级营养级的生物量并非全部被下一级取食，部分物质无法消化，以粪便形式损失。
Of the energy assimilated, a large proportion is lost as heat during respiration, used for metabolic processes such as movement, growth and maintenance.
在被同化的能量中，大部分在呼吸作用中以热能形式散失，用于运动、生长和维持等代谢过程。
Some energy is also lost through excretion of nitrogenous wastes and in plants, through energy used in transpiration and support.
部分能量还通过排泄含氮废物损失；植物中，蒸腾作用和支撑结构也消耗能量。
Typically, only about 10% of the energy at one trophic level is transferred to the next, resulting in a pyramid shape of energy. This limits the length of food chains.
通常只有约 10% 的能量从一个营养级传递到下一级，形成能量金字塔。这限制了食物链的长度。
9. Cardiac Cycle & ECG Interpretation | 心动周期与心电图解读
Question: Interpret the key features of a normal electrocardiogram (ECG) trace and explain their physiological causes. (5 marks)
题目：解读正常心电图 (ECG) 的关键波形特征，并解释其生理成因。(5分)
Answer & Commentary:
The P wave represents atrial depolarisation, triggered by the sinoatrial node (SAN). This electrical activity leads to atrial contraction (systole), pushing blood into the ventricles.
P 波代表心房除极，由窦房结 (SAN) 触发。该电活动导致心房收缩（心房收缩期），将血液挤入心室。
The QRS complex indicates ventricular depolarisation, which spreads from the atrioventricular node (AVN) through the bundle of His and Purkinje fibres. Ventricular contraction pumps blood out of the heart.
QRS 波群表示心室除极，电信号从房室结 (AVN) 经希氏束和浦肯野纤维传导。心室收缩将血液泵出心脏。
The T wave corresponds to ventricular repolarisation — the ventricles return to a resting electrical state and relax (diastole). Atrial repolarisation is hidden within the QRS complex.
T 波对应心室复极——心室恢复静息电位并舒张（舒张期）。心房复极波掩埋在 QRS 波群中，通常不易见。
A normal PR interval (0.12–0.20 s) reflects the delay at the AVN, allowing complete ventricular filling. Abnormalities in the ECG, such as an elevated ST segment, can indicate myocardial infarction.
正常的 PR 间期 (0.12–0.20 s) 代表房室结的延迟，确保心室充分充盈。ECG 异常，如 ST 段抬高，可能提示心肌梗死。
10. Humoral Immunity & Vaccination | 体液免疫与疫苗接种
Question: Describe how B lymphocytes are activated and explain why a second exposure to a pathogen results in a faster and stronger immune response. (6 marks)
题目：描述 B 淋巴细胞的激活过程，并解释为何第二次接触同一病原体会引发更快、更强的免疫应答。(6分)
Answer & Commentary:
B cells have specific antibody receptors on their surface. When a pathogen with complementary antigens enters the body, the B cell binds the antigen and internalises it through receptor‑mediated endocytosis.
B 细胞表面带有特异性抗体受体。当携带互补抗原的病原体进入体内，B 细胞与之结合，并通过受体介导的胞吞作用内化抗原。
The B cell processes the antigen and presents antigen fragments on its MHC class II molecules. An activated helper T cell with the complementary T‑cell receptor recognises this complex and releases cytokines that stimulate the B cell.
B 细胞加工抗原，并将抗原片段呈递在其 MHC II 类分子上。已活化的辅助 T 细胞通过互补受体识别此复合物，并释放细胞因子刺激 B 细胞。
The activated B cell proliferates and differentiates into plasma cells that secrete large quantities of antibodies, and memory B cells that persist long term.
活化的 B 细胞增殖并分化为浆细胞和记忆 B 细胞。浆细胞大量分泌抗体，记忆 B 细胞则长期在体内存活。
Upon second exposure, memory B cells recognise the antigen rapidly and differentiate into plasma cells faster and in greater numbers, producing a stronger antibody response before symptoms develop. This is the basis of immunological memory and vaccination.
再次暴露时，记忆 B 细胞迅速识别抗原，更快、更大量地分化为浆细胞，在症状出现前产生更强的抗体应答。这是免疫记忆和疫苗接种的原理。
Vaccines contain weakened or inactivated pathogens, or their antigens, triggering a primary immune response and generating memory cells without causing disease, providing long‑term protection.
疫苗含有减毒或灭活的病原体或其抗原，可激发初次免疫应答并产生记忆细胞，不会引发疾病，从而提供长期保护。
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